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Evaluate $$\lim_{x \to -\infty} \left(\frac{\sqrt{1+x^2}-x}{x} \right)$$

I tried by taking $x^2$ out of the root by taking it common.

i.e: $$\lim_{x \to -\infty} \left(\frac{x\sqrt{\frac{1}{x^2}+1}-x}{x} \right)$$ and then cancelling the x in numerator and denominator

$$\lim_{x \to -\infty} \left(\frac{\sqrt{\frac{1}{x^2}+1}-1}{1} \right)$$

then substituting $x= -\infty$ in the equation, we get, $$\lim_{x \to -\infty} \left(\frac{\sqrt{0+1}-1}{1} \right)$$ which equals to $0$. But it is not the correct answer. What have I done wrong.

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    $\begingroup$ When $x$ is close to $-\infty$, $x<0$ so $\sqrt{x^2}=|x|=-x$. This is a classic mistake: you did it once, you will never do it again :) $\endgroup$ – Taladris Aug 3 '14 at 12:46
  • $\begingroup$ But how does it matter $\sqrt{1-0} = |1| = 1$. I'm not sure. Correct me if i'm wrong. $\endgroup$ – TESLA____ Aug 3 '14 at 12:50
  • $\begingroup$ Ahh I think I see, well what about the x term following the root! does it go to -1 as you have suggested :) ? $\endgroup$ – Just_a_fool Aug 3 '14 at 12:58
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When $x\lt 0$, we have $$\sqrt{1+x^2}\not =x\sqrt{\frac{1}{x^2}+1}.$$ (Note that LHS is positive and that RHS is negative!)

You can set $-x=t\gt 0$. $$\lim_{x\to -\infty}\frac{\sqrt{1+x^2}-x}{x}=\lim_{t\to\infty}\frac{\sqrt{1+(-t)^2}+t}{-t}=\lim_{t\to\infty}\left(-\sqrt{\frac{1+t^2}{t^2}}-1\right)=-2.$$

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Hint: Multiply numerator and denominator by $\sqrt{1+x^2} + x$.

$$\lim_{x \to -\infty} \left(\frac{\sqrt{1+x^2}-x}{x} \right)\cdot \frac{\sqrt{1+x^2} +x}{\sqrt{1 + x^2}+x} = \lim_{x\to -\infty}\frac{1 + x^2 - x^2}{x(\sqrt{1 + x^2} + x)}$$

$$= \lim_{x\to -\infty}\frac 1{x\sqrt{1 + x^2} + x^2}$$

Can take it from here?

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    $\begingroup$ I don't think this is a good method, in this particular case, because the denominator is still an indeterminate form. $\endgroup$ – egreg Aug 3 '14 at 13:03
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    $\begingroup$ It's still a good method, just it should be applied one more time. $\endgroup$ – pointer Aug 3 '14 at 13:26
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Note that $\tan t$ maps $(-\pi/2,\pi/2)$ to $(-\infty,\infty)$ and is increasing on this interval. Consequently we can validly substitute $x=\tan t$ and get $$\lim_{x \to -\infty} \left(\frac{\sqrt{1+x^2}-x}{x} \right)=\lim_{t \to -\frac{\pi}{2}^+} \left(\frac{\sqrt{1+\tan^2(t)}-\tan t}{\tan t} \right)$$ which is readily computed if we recall some trigonometry. (Note that we must ensure the correct sign of the square root.)

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  • $\begingroup$ Can you explain how it is valid to substitute $x = \tan{t}$? $\endgroup$ – TESLA____ Aug 3 '14 at 13:07
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    $\begingroup$ See the first line: $\tan t$ can assume any real value, so we can assume WLOG that $x$ is the tangent of some angle $t$. $\endgroup$ – Deathkamp Drone Aug 3 '14 at 13:30
  • $\begingroup$ @Semiclassical, I was tempted to use $x=\tan2y$ before seeing this $\endgroup$ – lab bhattacharjee Aug 3 '14 at 13:50
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Setting $h=-\dfrac1x,$

$$\lim_{x \to -\infty} \left(\frac{\sqrt{1+x^2}-x}{x} \right)$$

$$=\lim_{h \to 0^+}\left(\frac{\sqrt{1+\dfrac1{h^2}}+\dfrac1h}{-\dfrac1h} \right)$$

$$=-\lim_{h \to 0^+}\left(\dfrac{\sqrt{h^2+1}}{|h|}+\dfrac1h \right)h$$

$$=-\lim_{h \to 0^+}\left(\sqrt{h^2+1}+1 \right)\text{ as }h\ne0\text{ as }h\to0^+$$

$$=-\left(\sqrt{0^2+1}+1 \right)$$

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  • $\begingroup$ @TESLA____, How about this? $\endgroup$ – lab bhattacharjee Aug 3 '14 at 13:32
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$$ \displaylines{ \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {1 + x^2 } - x}}{x} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)x^2 } - x}}{x} \cr = \mathop {\lim }\limits_{x \to - \infty } \frac{{\left| x \right|\sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)} - x}}{x} \cr = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - x\sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)} - x}}{x} \cr = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - x\left( {\sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)} + 1} \right)}}{x} = -\mathop {\lim }\limits_{x \to - \infty } \sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)} + 1 =- 2 \cr} $$

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